Metamath Proof Explorer


Theorem ssopab2

Description: Equivalence of ordered pair abstraction subclass and implication. (Contributed by NM, 27-Dec-1996) (Revised by Mario Carneiro, 19-May-2013)

Ref Expression
Assertion ssopab2
|- ( A. x A. y ( ph -> ps ) -> { <. x , y >. | ph } C_ { <. x , y >. | ps } )

Proof

Step Hyp Ref Expression
1 id
 |-  ( ( ph -> ps ) -> ( ph -> ps ) )
2 1 anim2d
 |-  ( ( ph -> ps ) -> ( ( z = <. x , y >. /\ ph ) -> ( z = <. x , y >. /\ ps ) ) )
3 2 aleximi
 |-  ( A. y ( ph -> ps ) -> ( E. y ( z = <. x , y >. /\ ph ) -> E. y ( z = <. x , y >. /\ ps ) ) )
4 3 aleximi
 |-  ( A. x A. y ( ph -> ps ) -> ( E. x E. y ( z = <. x , y >. /\ ph ) -> E. x E. y ( z = <. x , y >. /\ ps ) ) )
5 4 ss2abdv
 |-  ( A. x A. y ( ph -> ps ) -> { z | E. x E. y ( z = <. x , y >. /\ ph ) } C_ { z | E. x E. y ( z = <. x , y >. /\ ps ) } )
6 df-opab
 |-  { <. x , y >. | ph } = { z | E. x E. y ( z = <. x , y >. /\ ph ) }
7 df-opab
 |-  { <. x , y >. | ps } = { z | E. x E. y ( z = <. x , y >. /\ ps ) }
8 5 6 7 3sstr4g
 |-  ( A. x A. y ( ph -> ps ) -> { <. x , y >. | ph } C_ { <. x , y >. | ps } )